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2 years ago

What is the mass of 6 atom(s) of copper in grams?

Chemistry

1 answer:

USPshnik [31]2 years ago

4 0

Answer:

6.33×10¯²² g

Explanation:

From the question given above, the following data were obtained:

Number of atoms = 6 atoms

Mass of copper (Cu) =?

From Avogadro's hypothesis, we understood that:

6.02×10²³ atoms = 1 mole of Cu

But 1 mole of Cu = 63.5 g

Thus,

6.02×10²³ atoms = 63.5 g of Cu

Finally, we shall determine the mass of 6 atoms of copper. This can be obtained as illustrated below:

6.02×10²³ atoms = 63.5 g of Cu

Therefore,

6 atoms = (6 × 63.5) / 6.02×10²³

6 atoms = 6.33×10¯²² g of Cu

Therefore, the mass of 6 atoms of copper is 6.33×10¯²² g.

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Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

3 years ago

The mineral magnesite contains magnesium carbonate, MgCO3 (molar mass= 84 g/mol), and other impurities. When a 1.26-g sample of

Leviafan [203]

The answer is 33.33 %

The explanation:

According to the reaction equation:

MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)

we can see that 1 mole of MCO3 will produce → 1 mole of CO2

-Now we need o get number of mole of CO2:

and when we have 0.22 g of CO2, so number of mole = mass / molar mass

moles = 0.22 g / 44 g/mol = 0.005 mole

∴ moles of Mg = moles of CO2 = 0.005 mole

∴ mass of Mg = moles * molar mass

= 0.005 * 84 /mol = 0.42 g

∴ Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100

= 33.33 %

6 0

3 years ago

Read 2 more answers

The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the

Sonbull [250]

Answer:

The rate of consumption of $NH_{3}$ is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

where $-\frac{\Delta [NH_{3}]}{\Delta t}$ represents rate of consumption of $NH_{3}$ , $-\frac{\Delta [O_{2}]}{\Delta t}$ represents rate of consumption of $O_{2}$ , $\frac{\Delta [N_{2}]}{\Delta t}$ represents rate of formation of $N_{2}$ and $\frac{\Delta [H_{2}O]}{\Delta t}$ represents rate of formation of $H_{2}O$ .